IST module 4

Module IV
Laplace TransformLaplace Transform
Vijaya Laxmi., Dept. of EEE, BIT, Mesra 1

• The ordinary differential equations can be solved by following these
three steps:
 Determination of complementary function
 Determination of particular integral
 Determination of arbitrary constants Determination of arbitrary constants
• The solution is directly obtained in time domain as the process of
solving differential equation deals with functions at every step.
• The classical method is the last option for solving the differential
equation , when transformation methods fail.
• The classical methods are difficult to apply when excitation function
consists of derivatives, and transform methods are proved to be
superior.

• Transformation is somewhat similar to logarithmic operation.
• To find the product or quotient of two numbers:
 Obtain the logarithm of two numbers
 Add or subtract
 Take antilogarithm to get the product or quotient.
• In similar way, integro-differential equation by Laplace transform
method
 Transform the time-domain integro-differential eqn. into an
algebraic eqn. in s
 Find the roots of the characteristic polynomial
 Find the inverse Laplace transform to obtain the solution in time-
domain.

Comparison between logarithmic and
Laplace transform

Advantages of Laplace transform
• The solution of differential equations is a systematic
procedure.
• Initial conditions are automatically considered in a specified
transformation.
• It gives the complete solution, both complementary as well as• It gives the complete solution, both complementary as well as
particular integral in one operation.

Laplace transformation
• The Fourier transform is applicable to a large variety of signals
and widely used tool in engineering. But, it is applicable to
only those signals which satisfy the condition
  )1...(

dttf
• Therefore, many time functions which are of interest in
engineering cannot be handled by this method such as ramp,
parabolic etc, as the integral is not converging and the
functions are not Fourier transformable.
• In order to handle these types of functions, a convergence
factor is introduced to modify the transformation.
  )1...(
dttf
econvergencabsoluteensuretoenoughelandnumberrealiswheree t
arg, 

• The new transformation is
• Eqn. (2) now becomes
    )2...(
0



 dteetfsF tjt 
Lower limit is taken as 0 instead of -∞, since for σ>0, convergence factor will
Diverge when t->-∞ . Hence, the transforma on in eq. (2) ignores all
information contained in f(t) prior to t=0.
• Eqn. (2) now becomes
• The condition for Laplace transform to exist is
    )3...(
0



 dtetfsF st
 jswhere ,
F(s) is called the Laplace transform of f(t) and denoted by
F(s)=ʆ[f(t)] …(4)
  )5...(
0



dtetf st

• f(t) and F(s) are called the Laplace transform pair.
• Laplace transform thus changes time-domain to frequency-
domain and inverse Laplace transform changes form
frequency domain to time domain as given below.
       )6...(
2
11




j
j
st
dsesF
j
sFtf L



Theorems
• Theorem 1: The Laplace transform of the sum of two
functions is equal to the sum of the Laplace
transforms of the individual functions.
•           dtetftftftfL st
 

•          
   
     
    )7...(21
21
0
2
0
1
0
2121
sFsF
tftf
dtetfdtetf
dtetftftftf
LL
L
stst
st












• Theorem 2: The Laplace transform of a constant
times a function is equal to the constant times the
Laplace transform of the function.
      dtetkftkf st
L  


     
    )8...(
0
0
skFdtetfk
dtetkftkf
st
L






         )9...(
,
22112211 sFksFktfktfk
generalIn
L 

Laplace transform of important
functions
• Exponential function:   at
etf 
   
dtedteee tasstatat
L  




 
)10...(
1
00
as 



• Unit step function:  


 

otherwise
tfor
tU
0
01
       )11...(
11
0
0
s
e
s
dtetUtU stst
L 



0
ss
   )12...(
,
s
k
tU
Similarly
L 

• Sine and Cosine function:
     
)13...(
11
2
1
2
1


 




 
jsjsj
ee
j
tSin tjtj
LL
)13...(22


s
     
)14...(
11
2
1
2
1
22


 





 
s
s
jsjs
eetCos tjtj
LL

• Hyperbolic Sine and Cosine function:
 
  )16...(
2
1
)15....(
2
1
atat
atat
eeCoshat
eeSinhat




     
     )17...(
11
2
1
2
1
2
1
2
1
22
000
as
a
asas
ee
dteedteedteeeSinhat
atat
statstatstatat
LL
L









 







  )18...(, 22
as
s
CoshatSimilarly L


2

• Damped sine and cosine function:
     
 
   
 11
2
1
2
1

 




 
jasjasj
ee
j
tSine tjatjaat
LL
   
 
)19...(22




as
     
 
   
 
 
)20...(
11
2
1
2
1
,
22


 







 
as
as
jasjas
eetCose
Similarly
tjatjaat
LL

• Damped hyperbolic sine and cosine function:
     
     
 
 
)21...(
11
2
1
2
1
22
bas
as
basbas
eeCoshbte tbatbaat
LL






 
  22
bas 
     
     
 
 
)22...(
11
2
1
2
1
,
22
bas
b
basbasj
ee
j
Sinhbte
Similarly
tbatbaat
LL





 
The results can also be obtained using shifting property.

Problem
• Find
• Solution:






352
4
2
1
ss
s
L
2
3
5
1
6
2
1
3
4
352
4
s
s
ss
s















 
 2/3
11
2
1
2
56
2
1
2
3
5
1
6
2
1
352
4
2
312
2
3
12
352
tt
ee
ssss
s
ss
ss
ss
LLL





























































2222
22
2
2
4
1
4
5
4
11
4
1
4
5
4
5
4
1
4
5
4
2
1
2
3
2
5
2
4
352
4
,
ss
s
s
s
ss
s
ss
s
elyAlternativ
 















































































    
 2/3
4/4/4/4/
4/5
4/54/5
22
1
22
1
2
1
56
2
1
2
11
22
1
4/114/
2
1
4
1
4
5
4
1
11
4
1
4
5
4
5
2
1
352
4
tt
tttt
t
tt
ee
eeee
e
tSinhetCoshe
ss
s
ss
s
LLL











 













































































• The Laplace transform of e-at times a function is equal to the
Laplace transform of that function, with s replaced by (s+a).
• Proof:          
  )23...(
00
asF
dtetfdtetfetfe tasstatat
L

 




For damped sinusoid and hyperbolic functions also this can be used.For damped sinusoid and hyperbolic functions also this can be used.
   
   
 
   
 
)24...(
)24...(,
,
22
22
2222





















as
as
asFtCoseand
as
asFtSineTherefore
s
s
tCos
s
tSin
at
at
L
L
LL
   
    2222
,,
bas
as
Coshbte
bas
b
Sinhbteand atat
LL




 

  



0
dtett stnn
L
1
,
,
,







st
stn
st
n
s
e
dtevandntduThen
dtedvand
tupartsbygIntegratin
   
     
      1
0
11
0
1
0
1
0
0
0
0
|1|12
.....
21
1

















nn
nnstn
stnst
n
n
s
n
ss
n
t
sss
n
s
n
s
n
t
s
n
s
n
t
s
n
dtet
s
n
dtet
s
n
e
s
t
vduuvudvt
s
L
LL
L
    ...,
2|
,
1
, 3
2
2
s
t
s
tHence LL 

  
btSinAe at
L
    
 2222
bs
sSin
bs
bCos
A
CosbtSinSinbtCosAbtASin LL







    
    
 
 
 
 
 
 22
2222
bas
SinasbCos
A
bas
Sinas
bas
bCos
AbtSinAe
asFtfeAs
at
at
L
L















• Shifting Theorem:
• Proof:
    
       )29...(, 0
00 sFettUttfThen
sFtfIf
st
L
L



            dtettfdtettUttfttUttf
t
stst
L
0
0
0
0000




 
   
   sFedefe
def
stsst
ts
00
0
0
0













   
 












t
ttits
ddtttttand
ttttf
tt
ttUttfwhere
0,lim
,,,
;
;0
,
0
00
00
0
00

• Therefore, the four following functions are different:
 
   
   
   00
0
0
0
)(
)(
)(
)(
ttUttfiv
ttUtfiii
tUttfii
ttfi




00
 
   
       
     
       0000
00
00
00
)(
)(
)(
)(
,
ttUttSinttUttfiv
tttUSinttUtfiii
tUttSintUttfii
ttSinttfi
thentSintfIf











    22





s
sFandtSintfIf
    000)(





tSin
s
tCos
ttSinCosttCosSinttSini LL
22
00
022022














s
tsSintCos
tSin
s
s
tCos
s

       
00
00)(





tsSintCos
ttSintUttSinii LL
22
00





s
tsSintCos

    
   
 











 0
0
00
0
2
1
;1
;0
,;)( 

dtee
j
tt
tt
ttUasdtteSintttUSiniii
tjstjs
t
st
L
 
   























22
000
00
0
2
1
2




s
tsSintCos
e
js
e
js
e
j
dtee
j
st
tjstjs
t

      









22
00
0
0
sin)(



s
e
theoremshiftinggutSinettUttSiniii
st
st
LL
 s

Differentiation theorem
• If a function f(t) and its derivatives are both Laplace
transformable, and if L[f(t)]=F(s), then
• Proof:
     



0fssF
dt
tdf
L
      

• Proof:
      



0
, dtetftfLsFdefinitionBy st
   
stst
e
s
vdtedvand
dt
dt
tdf
dutfuLetpartsbygIntegratin









1
,
,,

         





















dt
tdf
ss
f
dte
dt
tdf
s
etf
s
sF
vduuvudvHence
Lstst 1011
,
00
0
0
0
It can be extended to higher order derivatives , when they are
Laplace transformable.
     



0, fssF
dt
tdf
Therefore L
Laplace transformable.
       
            

























0000 '2'
0
2
2
fsfsFsffssFs
dt
tdf
dt
tdf
s
dt
tdf
dt
d
dt
tfd
t
LLL
         
   
 
   
 









n
k
kknn
nnnnn
n
n
fssFs
fsffsfssFs
dt
tfd
generalIn L
1
1
1221
0
00...00,
         





0''0'0, 23
3
3
fsffssFs
dt
tfd
Similarly L

Integration theorem
• The Laplace transform of the first integral of a function f(t)
with respect to time is the Laplace transform of f(t) divided by
s, i.e.,
• Proof:
   
s
sF
dttf
t
L 





0


• Proof:
    














0 00
, dtedttfdttfdefinitionBy st
tt
L
   
stst
t
e
s
vdtedv
dttfdudttfuLet


 
1
0

 
   
    
s
sF
tf
s
dtetf
s
dttf
s
e
vduuvudvdttfpartsbygIntegratin
L
L
st
tst
t























1
00
1
,
000
0
0
00
   t t t
sF
dtdtdttfgeneralIn
n
L 



 
1 2
,...,,...,    
nn
s
sF
dtdtdttfgeneralIn L 








 0 0 0
21 ,...,,...,
Laplace transform of the indefinite integral of a function may be obtained as
      
 01
0
fdttfdttf
t
           
s
f
s
sF
fdttfdttfHence LLL
t











0
0,
1
1
0
f-1(0+) is the value of integral of
f(t) as t approaches to 0 from
positive side.

Multiplication by t
• Differentiation by s in the complex frequency domain
corresponds to multiplication by t in the time domain, i.e.,
• Proof:
    
ds
sdF
ttfL 
• Proof:
        ttfdtettfdte
ds
d
tf
ds
sdF
Lstst
 




00

Problem
• If
• Solution:
   
as
sFetf at

  1
,
 at
teFind L 
• Solution:
 
 2
11
asasds
d
te at
L










Problem
• Find
• Solution:
 tSintL 2
     
 
 222
22
222
2
2
2
22
32
1
















s
s
sds
d
tSin
ds
d
tSint LL

Division by t
        






s
dssF
t
tf
thensFtfIf LL ,
Proof:
    
 






 st
dsdtetfdssF     




ss
dsdtetfdssF
0
    
  















00
dt
t
e
tfdtdsetf
s
st
s
st
 
 













 


t
tf
dte
t
tf
L
st
0

Problem
 dstSin
t
tSin
s
LL 















t
tSin
Find L 
t s





























s
ss
ds
s
s
s






111
22
tantan
2
tan

Solution of linear differential equations
 
.)(tan10
012
2
functionexcitationtheisteandtsconsarebandbwhere
tewb
dt
dw
b
dt
wd

Taking Laplace transform of both sides
      tewb
dt
dw
b
dt
wd
LLLL 




01
2
2
Taking Laplace transform of both sides
  )0()( wssW
dt
dw
L  )0(')0()(
2
2
2
wswsWs
dt
wd
L 




    )()()0()()0(')0()(, 01
2
sEsWbwssWbwswsWsHence 

    )0(')0()()( 101
2
wwbssEsWbsbs 
Rearranging,
   )1....()0(')0()(
1
)( 12
wwbssE
bsbs
sW 

    )1....()0(')0()()( 1
01
2
wwbssE
bsbs
sW 


Taking inverse Laplace transform,
 
 








 
01
2
1 )0(')0()(
)()( 11
bsbs
wwbssE
sWtw LL
)()()(
)1(
0 sEsHsW
aswrittenbecanEquation

fnExcitationTotalXfnTransfertransformsponseor Re,

• Transfer function H(s) is the ratio of the transform of the
output (response) of that of the input (excitation) and is equal
to the reciprocal of the characteristic function.
• Total excitation function E0(s) consists of the Laplace
transform of the applied excitation, E(s) and contributions duetransform of the applied excitation, E(s) and contributions due
to the initial conditions, which do not depend on the applied
excitation.

Initial value theorem
• If the Laplace transform of f(t) is F(s), and the first derivative
of f(t) is Laplace transformable, then the initial value of f(t) is
• Proof:
      ssFLimtfLimf
st 

0
0
If the time limit exists
       














0
0
fssFdtetf
dt
d
tf
dt
d st
L
Let s approaches ∞Let s approaches ∞
      






  0
0
fssFLimdtetf
dt
d
Lim
s
st
s
s is not a function of t. By hypothesis, df(t)/dt being Laplace transformable,
the integral on the left side of the last equation exists. Therefore, it is
allowable to let s->∞ before integra ng. Then, LHS vanishes and we have,
    
    ssFLimfHence
fssFLim
s
s




0,
00
It is used for determining the initial
value of f(t) and its derivatives.

Final value theorem
• If the Laplace transform of f(t) is F(s), and if sF(s) is analytic on
the imaginary axis and the right half of s-plane, then,
• Proof:
   ssFLimtfLim
st 0

       














0
0
fssFdtetf
dt
d
tf
dt
d st
L
Let s approaches 0, thenLet s approaches 0, then
         







  0
0
0
0
0
fssFLimtfdtetf
dt
d
Lim
s
st
s
         
    ssFLimtfLimHence
fssFLimftfLimTherefore
st
st
0
0
,
00,




It is a very useful relation in the analysis and design of feedback control systems,
since it gives the final value of a time function by determining the behaviour of its
Laplace transform, when s tends to zero. However, this theorem is not valid if the
denominator of sF(s) contains any zero whose real part is zero or positive, which
is equivalent for analytic requirement of sF(s)Vijaya Laxmi., Dept. of EEE, BIT, Mesra 42

Gate function
• A rectangular pulse of unit height, starting at t=t1 and of
duration T is known as gate function and represented as
• Any function multiplied by a gate function will have zero value
outside the duration of the gate, (t1+T)<t<t1, and the value of
     TttUttUTGt  111
 ttU outside the duration of the gate, (t1+T)<t<t1, and the value of
the final function will be unaffected within the duration of the
gate, t1+T<t<t1+T.
• The equation of a gate function starting at the origin and of
origin will be
 1ttU 
     TtUtUTG 0
         TsTs
e
s
e
ss
TtUtUTGand LL 
 1
111
, 0

Problem: Sawtooth waveform
• The sawtooth waveform can be represented as
• Now,
        TtUtUt
T
E
TGt
T
E
tf 











 0
           TttU
T
E
ttU
T
E
TtUtUt
T
E
tf LLLL 


















    
     
 
  Ts
TsTsTsTs
eTs
Ts
E
Tsee
Ts
E
e
s
E
e
Ts
E
Ts
E
TtEUTtUTt
T
E
Ts
E
TtU
Ts
E
TTT
LL
TTt
T
E
L














11
1
2
222
2
2

• Alternatively,
       
       TtEUTtUTt
T
E
ttU
T
E
tftftftf

 321
TT
         
  Ts
TsTs
eTs
Ts
E
e
s
E
e
Ts
E
Ts
E
tftftftf LL





11
321
2
22

Problem: single half-sine wave
• The function can be represented as
• Now,
      
























2
2
2/
2
0
T
tUtUt
T
ASinTGt
T
ASintf

     






























2
22 T
ttU
T
SinAttU
T
SinAtf LLL 
22
,
T
tU
TT
tSin
T
ttUSinAs LL 


























 
       
 
2/
22
/2
/2
22
2
22
2
2222
,
Ts
e
Ts
T
T
tU
T
t
T
Sin
T
tU
T
t
T
Sin
tUt
T
SinttU
T
SinAs
LL
LL

















































  
   
 
 2/
22
2/
2222
1
/2
/2
/2
/2
/2
/2
,
Ts
Ts
e
Ts
T
A
e
Ts
T
A
Ts
T
AtfTherefore L
















• Alternatively,
     
     2/2/
22
21
TtUTt
T
ASintUt
T
ASin
tftftf



TT
    
 
 
 
 
 
 2/
22
2/
2222
1
/2
/2
/2
/2
/2
/2
Ts
Ts
e
Ts
TA
e
Ts
TA
Ts
TA
tfL
















Impulse function
• We have         atUtU
a
Limtft
a
a 

1
0

         1
0


atUtU
a
LimtFt
a
a LL
  1
...!3/2/111
11
3322
0
0





 











s
sasaas
a
Lim
s
e
sa
Lim
a
a
as
a

Periodic functions
• The Laplace transform of a periodic function with period T is
equal to 1/(1-e-Ts) times the Laplace transform of the first
cycle.
• Proof: Let f(t) be a periodic function of time period T. Let f1(t),
f2(t), … be the functions describing the first cycle, secondf2(t), … be the functions describing the first cycle, second
cycle etc., then
       
          ...22
...
111
321


TtUTtfTtUTtftf
tftftftf
      
 sF
e
sFeeetfL
Ts
TsTsTs
1
1
32
1
1
...1












Problem
• Determine the Laplace transform of the periodic, rectified
half-wave as shown

Solution
• The Laplace transform of single half-wave is
 
 
 
 2/
2
2
1 1
2
2
Ts
e
T
s
T
A
sF 





      
 
 
 
   
      
 
 2
2
2/2
22/2/
2/
2
2
2/
2
2
1
1
211
21
2
2
1
1
,
T
s
T
A
e
T
see
T
Ae
T
s
T
A
e
e
tfsFThen
Ts
TsTs
Ts
Ts
Ts
L














 













Problem
• Determine the Laplace transform of the periodic saw-tooth
waveform as shown

Solution
• The Laplace transform of waveform for first period is
• Hence,
        



  TsTs
ee
T
Ts
A
tfsF L 1
2
1
1
2
2
11
• Hence,
      
 
 
 


















 


s
Ts
Coth
T
Ts
A
se
eT
Ts
A
e
sF
tfsF Ts
Ts
Ts
L
1
22
2
1
1
1
2
2
1
1

Problem
• Determine the Laplace transform of the waveform f(t) as
shown

Solution
• The function f(t) can be written as the sum of step
function as
• Then,
           52442413  tUtUtUtUtUtf
• Then,
 
 ssss
ssss
eeee
s
s
e
s
e
s
e
s
e
s
sF
542
542
24431
1
24431





Problem
• The first derivative of a function f(t) is shown in Fig. by the
impulse train. Determine the Laplace transform of the
function f(t).

Solution
• Since, the above quantities are impulses, the Laplace
transform will be
   
sssssssss
eeeeeeeee
dt
tdf
sF L
98765432
'









eeeeeeeee 
       
   
s
sF
sFei
fgassussF
dt
tdf
sFLet L
'
.,.
00min,'








     
                   987654321,
'
, 11







 
tUtUtUtUtUtUtUtUtUtfTherefore
s
sF
sFtfHence LL

Problem
• Given a pulse f(t), find the transform F(s). Find L-1[F2(s)] to
get the triangular waveform as shown in Fig.

Solution
• The square wave can be represented as
• Squaring F(s), we get
     
   as
e
s
sFThen
atUtUtf



1
1
,
     
 
              atUatatUatttUsFLtf
ttU
s
As
ee
s
e
s
sF
L
asasas
222
1
,
21
1
1
1
21
1
2
1
2
2
2
2
2










Problem
• Find the Laplace transform of triangular wave.

Solution
• The triangular waveform can be represented as
 











212
10
00
tfort
tfort
tfor
tf


  20 tfor
       
2
2
2
1
1
00
21
2
s
ee
dtetdttedtetftfsF
ss
ststst
L






 

Applications of Laplace transform
Laplace transform can be used as mathematical tools for
determining the system response, subject to arbitrary input
functions. The steps involved are:
 The system must be represented in terms of differential or
integro-differential equation form.integro-differential equation form.
 Take LT of the system differential or integro-differential
equation together with input excitation to obtain an algebraic
equation in s, i.e., frequency domain.
 Obtain the inverse LT to get the solution in time domain.

Solution of linear differential equation
• Let a second order differential equation be
  )1...(012
2
tuya
dt
dy
a
dt
yd

Where a1, a0 are constants and u(t) is the
given excitation function
         tusUandtysYLet LL 
Taking Laplace transform on both sidesTaking Laplace transform on both sides
              
           
         
             )3...(
00
,
)2...(00
1
,
00,
000
01
2
.
111
.
1
01
2
.
101
2
01
.
2
asas
yyassU
sYtyor
yyassU
asas
sYor
yyassUsYasasor
sUsYayssYaysysYs
LL













Hence, the solution
not only depends on
the characteristic
roots but also on the
Excitation function and
initial conditions.

• Equation (2) can be rewritten as
      )4...(sUsHsY 
   transformexcitationTotalfunctionTransfertransformsponse Re
   tyoftransformLaplacetheissYWhere,
 
    
conditionsinitialtodue
onscontributiandtuLofconsistssUtransformexcitationTotal
inputoftransformLaplace
outputoftransformLaplace
sHfunctionTransfer
zerobetoconditionsinitialAll
,
, 

Heaviside partial fraction expansion
• The Laplace transformation is used to determine the solution
of integro-differential equation. The general form of
differential equation
• As a result of Laplace transform, transformed into an algebraic
equation in s which may be rearranged as
 tuxa
dt
dx
a
dt
xd
a
dt
xd
a nnn
n
n
n
 

11
1
10 ...
equation in s which may be rearranged as
   
 
 
nn
nnn
asasasasa
termsconditioninitialsU
sq
sp
sX





1
2
2
1
10 ...
Q(s)=0 is called the characteristic equation and the zeros of denominator
polynomial q(s) are roots.
x(t) is the inverse Laplace transform X(s), which is the ratio of two
polynomials. It becomes of a complex nature for which no direct Laplace
transform is available. In general, the transform expression X(s) must be
broken into simpler terms before any direct transform can be used.
 
 
 
 sq
sp
sBsBsBB
sq
sp
Then
nmifgeneralIn
nm
nm 


...,
,
2
110
1
The degree of p(s) is lesser
Or equal to that of q(s)

PFE: Simple zeros of q(s)
• If all the zeros of q(s) are simple,
   
 
 
     
n
sss
n
ss
K
ss
K
ss
K
ssssssss
sp
sq
sp
sX
n








...
...
21
321
21
   
 
   
 
 
    11211
1
1
...
,
,
1
1
ssssss
sp
sq
sp
ssKso
sq
sp
ssKwhere
nss
s
ss
js
j
j



















Example
• We have  
   321
35



sss
s
sX
 
321
3211





 
s
K
s
K
s
K
sX
321  sss
    
    
    
 
3
6
2
7
1
1
,
63
72
11
33
22
11













sss
sXhence
sXsK
sXsK
sXsK
s
s
s

PFE: Multiple zeros of q(s)
• If r of the n zeros of q(s) are alike, the function X(s) becomes,
   
 
 
     
   














ss
A
ss
A
ss
A
ss
K
ss
K
ss
K
ssssssss
sp
sq
sp
sX
r
i
i
i
i
i
i
n
sss
nn
rn
........................
...
2
21
121
2121
   


zerosrepeatedoftermsrzerossimpleoftermsrn
ssssssssssss iiin
)(
21
   
       
 
     
   
   
  ii
ii
ss
r
ir
r
i
ss
r
iri
ss
r
iri
ss
r
iir
sq
sp
ss
ds
d
r
A
sq
sp
ss
ds
d
A
sq
sp
ss
ds
d
A
sq
sp
ssA




































1
1
12
2
2
1
!1
1
,...,
!2
1
,

Example
• Let,
• Then,
   
     21
1
3


ssssq
sp
sX
 
       3
13
2
121120
1112 






 
s
A
s
A
s
A
s
K
s
K
sX
  
2
1
00  sssXK   
    
    
    
     11
!2
1
01
11
2
1
2
2
1
3
2
2
11
1
3
12
1
3
13
21
00









s
s
s
s
s
sXs
ds
d
A
sXs
ds
d
A
sXsA
sXsK
 
   3
1
1
1
1
22
1
2
1






ssss
sX

PFE: Complex conjugate zeros
• Suppose q(s) contains a pair of complex zeros,
• Since these zeros are distinct , the corresponding coefficients
are
 jsandjs  21
are
   
 
1112
1
1
**
1
ssss
ss
s
KofconjugatecomplexisKwhereKKand
sq
sp
ssK










Example
• Consider,
• Then,
 
84
1
2


ss
sX
 
     
   *
1
*
1
1
1
2222
2222
1
22
1
444
1
84
1
ss
K
ss
K
jsjs
jsssss
sX












• Therefore,
   112222 ssssjsjs 
    4/4/
22,22,
*
111
*
11
1
jKandjsXssK
jsjswhere
ss



 
   
   
tSin
e
e
j
e
j
js
j
js
j
sXtx
js
j
js
j
sX
t
tjtj
LLL
2
2
44
22
4/
22
4/
22
4/
22
4/
2
2222
111















 












• Alternatively,
 
   
 













sss
sX
att 12
222
22
2
2
1
84
1

 
  







 
Sinate
as
LAstSinetx att
22
12
25.0



Problem
• Solve
• Taking Laplace transform,
• Substituting the values,
    10,00,0672
....
 xxxxx
            0607702022
.
2
 sXxssXxsxsXs
   
 tt
ee
ssss
sXtx LLL
25.1
1
2
11
2
1
2
1
5.1
1
2
1
672
2














Problem
• Solve,
    30,00063
....
 xxwithxxx
            0603300
.
2
 sXxssXxsxsXs
 
   
  













tSinetxhence
sss
sX
t
2
2/15
5
32
,
2/155.1
2/15
5
32
63
3
5.1
222

Problem
• Given the set of simultaneous equations, with all initial
condition as zero, find x1(t) and x2(t).
 
 txxxxand
tUxxxx
53
5742
22
.
1
.
1
22
.
1
.
1


• Using Cramer’s rule,
 
  



















5
/5
31
742
1
1 s
sX
sX
ss
ss
 
 
 
        




















22222
2
2
1
21
2
14
21
183
52
3683
52
15305
35
7/51
ss
s
sss
s
s
sss
ss
s
ss
sX
  tSinetCosetx tt
2142831



• Similarly,
 
   
   
 
        

























222222
22
21
2
3
21
1
11
1
21
231111
52
17111
51
/5421
ss
s
ss
s
s
ss
s
ss
ss
sX
            


 



 
222222
212121 sssss
   tSinetCosetx tt
2321112



Kirchoff’s Laws
• Resistive element:
       
       sV
R
sItv
R
tiKCL
sRIsVtRitvKVL
RRRR
RRRR














11
:
:

• Capacitive element:
     
 
dt
dv
CtiKCL
vdtti
C
tvKVL
c
c
c
t
cc

 
:
0
1
:
0
     
     


0:
01
:
,
ccc
c
cc
CvssCVsIKCL
s
v
sI
sC
sVKVL
LTTaking

• Inductive element:
 
     

 0
1
:
:
0
L
t
LL
L
L
idttv
L
tiKCL
dt
di
LtvKVL
     
     
s
i
sL
sV
sIKCL
LissLIsVKVL
LTTaking
LL
L
LLc



0
:
0:
,

• Other elements:
  di
M
di
Ltv 21
 ,LTTaking 
 
dt
di
M
dt
di
Ltv
dt
di
M
dt
di
Ltv
12
22
21
11


         
         

00
00
,
1122222
2211111
MissMIiLsIsLsV
MissMIiLsIsLsV
LTTaking
 
 
     
     



















00
00
1222
2111
1
1
2
1
MiiLsV
MiiLsV
sI
sI
sLsM
sMsL

Problem
• Consider the circuit shown in the figure. The switch is thrown
from position 1 to 2 at time t=0. Just before the switch is
thrown, the initial conditions are iL(0+)=2A, vc(0+)=2V. Find
the current i(t) after the switch thrown. Assume L=1H, R=3Ω,
C=0.5F, V1=5V.C=0.5F, V1=5V.

Solution
• At t>0, by KVL,
 
         
 
ss
v
Cs
sI
LisLsIsRILTTaking
Vvidt
Cdt
di
LRi
c
c
t
50
0,
0
1
1
0






• Therefore,
     
 
ss
sI
s
s
s
v
Li
s
sI
Cs
LSR
ssCs
c
2
2
52
3
0
0
51














   
  
tt
ee
ssss
s
sIti LLLL
2
1111
2
1
1
1
21
32











Problem
• Consider the circuit shown in the figure. The switch is opened
at time t=0. Find the node voltages. The initial conditions are
iL(0+)=1A, vc(0+)=1V. Assume L=0.5H, G=1 mho, C=1F, V=1V.

Solution
• The transformed circuit for t>0 is shown. The node equations
can be written as,
       
          
 
s
i
CvsCsV
sL
sVsVor
s
i
CvsIsINode
L
c
L
cba






0
0
1
,
0
0:1
121
     
        
s
i
GsV
sL
sVsVor
s
i
sIsINode
L
L
ca




01
,
0
:2
212
         
s
i
sI
s
i
CvsILet LL
c



0
,
0
0, 21
 
 
 
 
   
 
   
 
 int
11
2
11
1
,
11
11
2
1
2
1
2
2
1
1
1
1
2
1
2
1
SCoste
s
s
sVtv
Coste
s
s
sVtvSolving
sI
sI
sV
sV
sL
G
sL
sLsL
sC
t
t
LL
LL



































Problem
• Derive the expression of i(t) in the circuit shown in figure
using Laplace transform.

Solution
• Applying KVL in the circuit, in time domain,
• Taking LT,
   tvvidt
Cdt
di
LRi c
t
 
 0
1
0
           
     v
LisV
sV
s
v
sI
sC
LissLIsRI
c
c
L
0
0
01
0






 
     
sC
sLR
s
v
LisV
sI
c
L
1
0
0




     
 
      
 
 
   tUee
ssL
ti
LCL
R
L
R
swhere
ssss
L
LCsLRs
L
sI
ssVtvvigAssu
tsts
cL
21
21
2
2,1
21
2
1
1
22
,
/1
/1/
/1
/1,1)(,000,min


















IST module 4

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